a: \(=\sqrt{5}-1\)

b: \(=\sqrt{2}-1\)

c: \(=\sqrt{3}+1\)

d: \(=\sqrt{13}+1\)

`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

`=1-sqrt2-sqrt2=1-2sqrt2`

a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)

b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)

c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)

d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)

\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)

\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)

a: Ta có: \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}-\sqrt{20}\)

\(=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}-2\sqrt{5}\)

\(=-\sqrt{7}-2\sqrt{5}\)

a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)

\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)

b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)

\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)

\(\Leftrightarrow B^3+9B-10=0\)

\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))

c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)

\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)

\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)

\(\Rightarrow C=1\)

d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)

\(=\sqrt[3]{3}+\sqrt[3]{2}\)

Vậy...

A = (2 + √2)/(1 + √2)

= √2(√2 + 1)/(1 + √2)

= √2

C = (2√3 - √6)/(√8 - 2)

= √6(√2 - 1)/[2(√2 - 1)]

= √6/2

E = (x√x + 1)/(√x + 1)

= (√x + 1)(x - √x + 1)/(√x + 1)

= x - √x + 1

A = $\frac{2 + \sqrt{2}}{1 + \sqrt{2}}$

Để rút gọn biểu thức này, ta nhân tử và chia tử cho $1 - \sqrt{2}$:

A = $\frac{(2 + \sqrt{2})(1 - \sqrt{2})}{(1 + \sqrt{2})(1 - \sqrt{2})}$

A = $\frac{-2\sqrt{2}}{-1}$

A = $2\sqrt{2}$

C = $\frac{2\sqrt{3} - \sqrt{6}}{\sqrt{8} - 2}$

Ta nhân tử và chia tử cho $\sqrt{2}$:

C = $\frac{(2\sqrt{3} - \sqrt{6})\sqrt{2}}{(\sqrt{8} - 2)\sqrt{2}}$

C = $\frac{4\sqrt{6} - 2\sqrt{3}}{2\sqrt{2}}$

C = $\frac{2\sqrt{6} - \sqrt{3}}{\sqrt{2}}$

Ta nhân tử và chia tử cho $\sqrt{6} + \sqrt{2}$:

C = $\frac{(2\sqrt{6} - \sqrt{3})(\sqrt{6} + \sqrt{2})}{(\sqrt{2})(\sqrt{6} + \sqrt{2})}$

C = $\frac{12 - 3\sqrt{2}}{2}$

C = $6 - \frac{3\sqrt{2}}{2}$

E = $\frac{x\sqrt{x+1}}{\sqrt{x+1}}$

E = $x\sqrt{\frac{x+1}{x+1}}$

E = $x$.

a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)

b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)

Câu a, bạn coi lại đề xem $a^2=6-3\sqrt{3}$ hay $a=6-3\sqrt{3}$???

b.

\(B=\frac{\sqrt{(x-2)+(x+2)+2\sqrt{(x-2)(x+2)}}}{\sqrt{x^2-4}+x+2}\)

\(=\frac{\sqrt{(\sqrt{x-2}+\sqrt{x+2})^2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}=\frac{1}{\sqrt{x+2}}\)

\(=\frac{1}{\sqrt{3+\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{(\sqrt{5}+1)^2}}=\frac{\sqrt{2}}{\sqrt{5}+1}\)

Nguyễn Hoàng trung: Chả qua nếu $a=6-3\sqrt{3}; b=2+\sqrt{3}$ thì kết quả sẽ đẹp hơn. Còn như đề thì vẫn rút gọn được.

\(A=\frac{a-\sqrt{ab}+b}{(\sqrt{a}-\sqrt{b})^2}=\frac{a-\sqrt{ab}+b}{a-2\sqrt{ab}+b}\)

\(2a^2=12-6\sqrt{3}=(3-\sqrt{3})^2\Rightarrow a=\frac{3-\sqrt{3}}{\sqrt{2}}\) (do $a\geq 0$)

\(2b^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\Rightarrow b=\frac{\sqrt{3}+1}{\sqrt{2}}\) (do $b\geq 0$)

\(\Rightarrow a+b=2\sqrt{2}; ab=\frac{\sqrt{3}(\sqrt{3}-1)(\sqrt{3}+1)}{2}=\sqrt{3}\)

Do đó: $A=\frac{2\sqrt{2}-\sqrt[4]{3}}{2\sqrt{2}-2\sqrt[4]{3}}$

\(A=\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}}{3-2}-\sqrt{3}=\sqrt{3}-\sqrt{2}-\sqrt{3}\)

\(=-\sqrt{2}\)